If `NaCl` is doped with `10^(-3) mol% GaCl_(3)`, what is the concentration of the cation vacancies?

Options

A. `1.2xx10^(19)`
B. `1.2xx10^(10)`
C. `1.2xx10^(6)`
D. `1.2xx10^(4)`

Explanation

100mol of NaCl are doped with `10^(-3)` mol of `GaCl_(3)`.
`therefore` 1 mol of NaCl is doped with `GaCl_(3)=(10^(-3))/(100)=10^(-5)`mol
As one `Ga^(3+)` ion is introduced , three `Na^(o+)` have to be removed to maintain the electrical neutraal So as one vacancy is filled by `Ga^(3+)`, two cation vacancies are formed.
`therefore` Concentrations of cation vacancy
`=2xx10^(-5)` mol/mol of NaCl
`=2xx10^(-5)xx6.023xx10^(23) "mol"^(-1)`
`=12.046xx10^(-18)"mol"^(-1)=1.2046xx10^(-19)"mol"^(-1)`